Probability of rolling a sum of $9$ with $6$ dice

Question

Like in topic, you have 6 dice. You sum their values. What is the probability you get 9? How do I calculate it?

Answer 1

General approach. There aren't that many ways to get $9$ from six dice. (I assume these are ordinary six-sided dice.) Up to reordering, there are only the following three combinations:

$$ 1, 1, 1, 1, 1, 4 $$

$$ 1, 1, 1, 1, 2, 3 $$

$$ 1, 1, 1, 2, 2, 2 $$

Calculate the number of ways to get each of these (including their reorderings), and divide by $6^6$ to obtain the probability. You should obtain the same value as Matthew Conroy in his answer, hopefully with a lot less computation. (Generating functions are a convenient way to obtain all of the probabilities at once, though.)

Answer 2

Let's assume the dice are six-sided and distinct (each of a different color, say). Then there are $6^6$ possible outcomes in the sample space. Let $x_k$ be the outcome on the $k$th die. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 9 \tag{1}$$ Equation 1 is an equation in the positive integers subject to the restriction that $x_k \leq 6$ for $1 \leq k \leq 6$. Notice that if $x_k > 6$ for some $k$, then the sum is at least $7 + 5 = 12 > 9$, so we do not need to worry about that restriction (as Brian Tung demonstrated in his answer).

A particular solution to equation 1 corresponds to the insertion of addition signs in five of the eight spaces between successive ones in a row of nine ones. For instance, $$1 1 + 1 + 1 1 + 1 + 1 + 1 1$$ corresponds to the outcome $x_1 = 2$, $x_2 = 1$, $x_3 = 2$, $x_4 = 1$, $x_5 = 1$, and $x_6 = 2$, while $$1 1 1 + 1 1 + 1 + 1 + 1 + 1$$ corresponds to the outcome $x_1 = 3$, $x_2 = 2$, $x_3 = x_4 = x_5 = x_6 = 1$. Hence, the number of solutions of equation 1 is the number of ways we can select five of the eight spaces between successive ones in a row of nine ones, which is $\binom{8}{5}$. Hence, the probability of obtaining a sum of $9$ when six six-sided dice are thrown is $$\frac{\binom{8}{5}}{6^6}$$